Question

find the Eigen values and Eigen Vectors of matrix
7 -2 0
-2 6 -2
0 -2 5​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Eigen values of the matrix are $\bf \lambda_1 = 3, \lambda_2 = 6,\lambda_3 = 9$ and their corresponding eigen vectors are

$\bf v_1 = (1,2,2), v_2 = (-2,-1,2),v_3 = (2,-2,1)$

Given:

• $\left[\begin{array}{ccc}7&-2&0\\-2&6&-2\\0&-2&5\end{array}\right]$

To find:

• Find the Eigen values and eigen vectors of matrix.

Solution:

Concept to be used:

Characteristics equation $[A-\lambda \: I]X=O$

Step 1:

Write the characteristics equation.

$\left[\begin{array}{ccc}7&-2&0\\-2&6&-2\\0&-2&5\end{array}\right] - \lambda\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

$\left[\begin{array}{ccc}7 - \lambda&-2&0\\-2&6 - \lambda&-2\\0&-2&5 - \lambda \end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

Determinant of characteristics equation equal to zero.

$\left |\begin{array}{ccc}7 - \lambda&-2&0\\-2&6 - \lambda&-2\\0&-2&5 - \lambda \end{array}\right | =0$

Expand the determinant along R1.

$(7 - \lambda) \{((6 - \lambda)(5 - \lambda) - 4) + 2( - 2(5 - \lambda) - 0) \} = 0$

$(7 - \lambda) (\lambda ^{2} - 11\lambda + 26) - 20 + 4\lambda = 0$

$(7 \lambda ^{2} - 77\lambda + 182 - \lambda ^{3} + 11 \lambda ^{2} - 26\lambda - 20 + 4\lambda = 0$

$\bf - {\lambda}^{3} + 18 {\lambda}^{2} - 99\lambda + 162 = 0$

Step 2:

Solve the characteristics equation and find eigen values.

Find the first root of equation by hit and trial method.

Put $\lambda = 3$

$- {(3)}^{3} + 18 {(3)}^{2} - 99(3) + 162 = 0$

$0 = 0$

Thus,

$\bf \lambda = 3$ is one root of the equation and $( \lambda - 3)$is one factor.

Find the other roots by dividing the cubic polynomial by the known factor.

$\lambda - 3 \: ) - {\lambda}^{3} + 18 {\lambda}^{2} - 99\lambda + 162 ( -{\lambda}^{2} + 15 {\lambda} - 54 \\ - {\lambda}^{3} + 3 {\lambda}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ( + ) \: \: \: \: ( - ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ 15{\lambda}^{2} - 99{\lambda} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ 15{\lambda}^{2} - 45{\lambda} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ( - ) \: \: \: \: \: \: ( + ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - \: \: \: \: \: \: \: \: \: \: \: \: \\ - 54{\lambda} + 162 \\ - 54{\lambda} + 162 \\ ( + ) \: \: \: \: ( - ) \\ - - - - - \\ 0 \\ - - - - -$

Thus,

The quotient polynomial is the other factor.

Factorise the $-{\lambda}^{2} + 15 {\lambda} - 54 = 0$ to find the other two factors.

$\bf \lambda = 9$

or

$\bf \lambda = 6$

Thus,

The eigen values of the matrix are

$\bf \lambda_1 = 3, \lambda_2 = 6,\lambda_3 = 9$

Step 3:

Find the eigen vector of the matrix.

$[A-\lambda \: I \: ]X=0$

Find the eigen vector for eigen value 3.

$\left[\begin{array}{ccc}7 - 3&-2&0\\-2&6 -3&-2\\0&-2&5 -3 \end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

$\left[\begin{array}{ccc}4&-2&0\\-2&3&-2\\0&-2&2 \end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

$4x - 2y = 0...eq1 \\ - 2x + 3y - 2z = 0...eq2 \\ - 2y + 2z = 0 ...eq3$

Add eq1 and eq3

$4x - 4y + 2z = 0 ...eq4$

use eq2 and 4 to find the values of x,y,z

$\frac{x}{6 - 8} = \frac{ - y}{ - 4 + 8} = \frac{z}{8 - 12}$

$\frac{x}{ - 2} = \frac{ - y}{ 4} = \frac{z}{ - 4}$

or

$\frac{x}{ 1} = \frac{ y}{ 2} = \frac{z}{ 2}$

Eigen vector for $\lambda_1 = 3$ is

$\left[\begin{array}{ccc}1\\2\\2\end{array}\right]$

According to the same method, we can calculate the eigen vectors of other two eigen values.

Eigen vector for $\lambda_2 = 6$ is

$\left[\begin{array}{ccc}-2\\-1\\2\end{array}\right]$

Eigen vector for $\lambda_3 = 9$ is

$\left[\begin{array}{ccc}2\\-2\\1\end{array}\right]$

Thus,

Eigen values of the matrix are $\bf \lambda_1 = 3, \lambda_2 = 6,\lambda_3 = 9$ and their corresponding eigen vectors are$\bf v_1 = (1,2,2), v_2 = (-2,-1,2),v_3 = (2,-2,1)$

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